At low temperatures,the slow addition of molecular bromine to $CH_2=CH-CH_2-C \equiv CH$ gives:

  • A
    $CH_2=CH-CH_2-CBr=CHBr$
  • B
    $BrCH_2-CHBr-CH_2-C \equiv CH$
  • C
    $CH_2=CH-CH_2-CH_2-CBr_3$
  • D
    $CH_3-CBr_2-CH_2-C \equiv CH$

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